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Ivenika [448]
3 years ago
12

Why do scientists need a standard system of measurement?

Physics
2 answers:
djyliett [7]3 years ago
7 0
The answer is so that everyone is measuring the same distance and counting that distance as the same everywhere. one example from real life why it is important to use standard units is the loss of the NASA mars orbiter. some people decided that the units were to be in metric and others thought that the units were in imperial (english) system. so the orbiter was lost.

let's eliminate some choices
A. to allow for consistientcy with measurement data:
not really, just make sure the instruments are accurate.

B. to use tools marked in customary units: this could be true because a custom is something that is accepted in a wide area and if it is standard, then it is customary so b is the answer

C. so they can keep measurement data secret: nope

D. to record with easier abbreviations: not really

the answer is B
yKpoI14uk [10]3 years ago
4 0


None of the given choices is worth much.

The scientific community ... ALL of them ... need a standard system
of measurement so that one scientist can understand the results of
another scientist's work. 

Science is not a bunch of recluse hermits each working alone in their
dusty dungeons.  Science is a  community.  Each scientist publishes
a very detailed description of what he tried and what results he got. 
If other scientists read it and get ideas of how they could chase
answers to the same questions, they can pick up where the first
scientist left off, build on his work, or go off in slightly different
directions. 

None of that can happen unless everyone precisely understands
everyone else's measurements.

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What forms of rocks are produced when rocks travel to fewer stages in the cycle?
Veronika [31]

Answer:

The two major sources of energy for the rock cycle are also shown; the sun provides energy for surface processes such as weathering, erosion, and transport, and the Earth's internal heat provides energy for processes like melting, and metamorphism

4 0
3 years ago
A 0.439 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.307 kg puck moving initially along th
lesya [120]

Answer:

u2 = 0.266 m/s

Explanation:

Let the left Puck mass at rest = m1 =0.307 Kg

mass of the right puck m2 = 0.439 kg

velocity of m1 before collision v1= 2.19 m/s

velocity of m2 before collision v2 = 0m/s

velocity of m1 after collision u1 =1.19 m/s

velocity of m2 after collision u2 = ? m/s

θ = 37°

<u>Solution:</u>

Before collision:

Momentum (y-axis ) before collision= 0 Kgm/s

Momentum (x-axis ) before collision= m1v1 + m2v2 = 0.307 Kg x 2.19 m/s + 0

= 0.672 Kgm/s

After collision:

Momentum (y-axis ) after collision= m1u1 sinθ  + m2u2 sinθ

= 0.307 x 1.19 m/s sin 37 °  + 0.439 x u2 sin 37°

= 0.22 + 0.26 u2

Momentum (x-axis ) after collision= m1u1 cosθ  + m2u2 cos θ

= 0.307 x 1.19 m/s cos 37 ° + 0.439 x u2 cos 37°

= 0.29 + 0.35 u2

According to law of conservation momentum

momentum before collision = momentum after collision.

0 + 0.672 Kgm/s =  0.22 Kgm/s  + 0.26 kg u2 + 0.29 Kgm/s + 0.35 kg u2

0.672 Kgm/s = 0.51 Kgm/s + 0.61 u2

u2 = 0.266 m/s

6 0
4 years ago
Why do atoms do group 1 element lose electrons to form cations
Mrac [35]

Answer:

when a element of 1 group take part in reaction, its atom looses outer electron and form positively charged ions called Cation.

Explanation:

8 0
3 years ago
A racing car whose mass is 1.2 X 10^3 kg is travelling at 8.9 m/s. It stops with a constant deceleration in a distance of 1.8X10
Alexeev081 [22]

given that initial speed of the car is

v_i = 8.9 m/s

now after travelling the distance d = 1.8 * 10^1 m the car will stop

so here we can use kinematics to find the acceleration of car

v_f^2 - v_i^2 = 2 a d

0 - 8.9^2 = 2 a d

here we have

- 79.21 = 2*(18)*a

a = -2.2 m/s^2

net force applied due to brakes of car is given by Newton's II law

F = ma

here we have

mass = 1.2 * 10^3 kg

F_{net} = 1.2 * 10^3 * 2.2

F_{net} = 2.64 * 10^3 N

now we can say

F_{net} = F_1 + F_2

2.64 * 10^3 = 1.8 * 10^3 + F_2

F_2 = 8.4 * 10^2 N

So the force applied due to brakes is given as above

7 0
3 years ago
An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindston
Dmitry_Shevchenko [17]

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

6 0
4 years ago
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